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SubjectsMathematics
समुच्चय और उनके निरूपण
In this Class 10 Mathematics topic from the chapter “Sets,” students learn how to describe and represent a collection of well-defined objects using clear mathematical language. They explore common forms such as descriptive statements, roster or tabular notation, and set-builder notation, while identifying elements and understanding the symbols used for membership and non-membership. The topic builds accuracy in reading, writing, comparing, and interpreting sets, providing a foundation for later ideas involving relationships and operations on sets.
Correct answer: A. U = {−1, 0, 1}
Explanation: Rewrite the equation as x³ − x = 0 and factor it: x(x² − 1) = 0, so x(x − 1)(x + 1) = 0. The possible solutions are x = 0, x = 1, and x = −1. All three are integers and therefore satisfy the restriction x ∈ ℤ. Hence the set is U = {−1, 0, 1}, making option A correct.
Correct answer: A. 8
Explanation: Because x is an integer and x² < 25, x must lie between −4 and 4 inclusive. Thus the possible values are −4, −3, −2, −1, 0, 1, 2, 3, and 4. The additional condition 1 ≤ x² excludes only x = 0, because 0² = 0. The remaining eight integers are −4, −3, −2, −1, 1, 2, 3, and 4. Hence option A is correct.
Correct answer: A. R = {1, 4, 9, 36}
Explanation: The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. We must retain only those divisors that are also perfect squares. Here, 1 = 1², 4 = 2², 9 = 3², and 36 = 6². The numbers 2, 3, 6, 12, and 18 are not perfect squares. Therefore, the required set is R = {1, 4, 9, 36}, so option A is correct.
Correct answer: A. {1, 5, 7}
Explanation: Examine the integers 1 through 10. The numbers divisible by 2 are 2, 4, 6, 8, and 10; the numbers divisible by 3 are 3, 6, and 9. Since the phrase “neither 2 nor 3” excludes every number in either group, remove their union. The numbers left are 1, 5, and 7. Hence the required members are {1, 5, 7}, making option A correct.
Correct answer: A. Z = {−2, 1}
Explanation: A product is zero when at least one of its factors is zero. Therefore, solve x − 1 = 0 or x + 2 = 0. The first equation gives x = 1, while the second gives x = −2. Both values are integers and satisfy the original equation: (1−1)(1+2)=0 and (−2−1)(−2+2)=0. Thus the roster form is {−2, 1}; order does not matter in a set, so A is correct.
Correct answer: A. A = {−4, −2, 0}
Explanation: First list the integers satisfying −5 ≤ x < 2: they are −5, −4, −3, −2, −1, 0, and 1. Now select the even integers from this list. The even numbers are −4, −2, and 0; negative integers can also be even because divisibility by 2 is what matters. The endpoints −5 and 2 are not included as even members: −5 is odd and 2 is outside the strict upper bound. Hence A = {−4, −2, 0}.
Correct answer: A. B = {4, 5, 6, 7, 8, 9}
Explanation: Since x is a natural number and x < 10, the possible values are 1 through 9. A two-digit square must be at least 10 and at most 99. The first natural number whose square is at least 10 is 4, because 3² = 9 but 4² = 16. Every value from 4 through 9 has a square between 16 and 81, so each qualifies. Therefore B = {4, 5, 6, 7, 8, 9}, making option A correct.
Correct answer: A. D = {2n : n ∈ Z, 0 ≤ n ≤ 4}
Explanation: Every element of D is twice an integer, and the required values of n are 0, 1, 2, 3, and 4. Thus 2n produces 0, 2, 4, 6, and 8 exactly when n ∈ Z and 0 ≤ n ≤ 4. Option B can depend on whether a convention includes zero in N, so option A is the unambiguous form.
Correct answer: A. E = {3, 8}
Explanation: The positive divisors of 10 are 1, 2, 5, and 10. Set x + 2 equal to each divisor: x = −1, 0, 3, and 8, respectively. With the usual school convention N = {1, 2, 3, ...}, only 3 and 8 belong to the natural numbers. Hence E = {3, 8}, making option A correct. Option C incorrectly includes 0, and option D lists divisors rather than the corresponding values of x.
Correct answer: A. G = {-1, 0, 1, 2, 3}
Explanation: Rearrange the inequality: x² − 2x − 8 < 0. Factoring gives (x − 4)(x + 2) < 0, which holds strictly when −2 < x < 4. Since x must be an integer, the possible values are −1, 0, 1, 2, and 3. The endpoints −2 and 4 give equality, not a strict inequality, so they are excluded. Thus option A is correct.
Correct answer: A. English: Y = {3} | हिन्दी: Y = {3}
Explanation: Rewrite the equation as x² + x − 12 = 0. Factoring gives (x + 4)(x − 3) = 0, so the algebraic solutions are x = −4 and x = 3. However, the set is restricted to natural numbers. Under the usual school convention, −4 is not a natural number, whereas 3 is natural. Consequently only 3 belongs to Y, and the required set is Y = {3}. Therefore option A is correct.
Correct answer: A. English: Z = {0, 1, 2, 3} | हिन्दी: Z = {0, 1, 2, 3}
Explanation: First solve x² ≤ 9. This gives −3 ≤ x ≤ 3, so the possible integers are −3, −2, −1, 0, 1, 2, and 3. Next solve x + 1 > 0, which gives x > −1. Among the possible integers, this retains 0, 1, 2, and 3; −1 is excluded because the inequality is strict. The intersection is therefore Z = {0, 1, 2, 3}, making option A correct.
Correct answer: A. D = {n³ : n ∈ ℕ, 1 ≤ n ≤ 4}
Explanation: The elements 1, 8, 27, and 64 are the cubes of 1, 2, 3, and 4 respectively: 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³. Therefore, the set can be described as D = {n³ : n ∈ ℕ and 1 ≤ n ≤ 4}. The other options generate squares, powers of 2, or multiples of 4, so they do not produce exactly the given set.
Correct answer: A. E = {1, 2, 3, 4, 5}
Explanation: Solve the defining inequality: 2x − 1 ≤ 9 gives 2x ≤ 10 and hence x ≤ 5. Since x belongs to the natural numbers, and natural numbers here are taken as 1, 2, 3, …, the possible values are 1, 2, 3, 4, and 5. Thus the roster form is E = {1, 2, 3, 4, 5}. The equality sign includes 5.
Correct answer: A. F = {−3, −2, −1, 1, 2, 3}
Explanation: The strict inequality −4 < x < 4 permits the integers −3, −2, −1, 0, 1, 2, and 3; the endpoints −4 and 4 are excluded. The additional condition x ≠ 0 removes 0 from this list. Therefore, F = {−3, −2, −1, 1, 2, 3}. Option C incorrectly retains zero, while option B incorrectly includes the endpoints.
Correct answer: B. P = {2, 4, 6, 8}
Explanation: Under the standard school convention used here, natural numbers begin with 1. The one-digit natural numbers are 1 through 9, and the even ones among them are 2, 4, 6, and 8. Zero is not included because it is not being counted as a natural number in this convention, and 10 is not one-digit. Therefore the correct roster form is P = {2, 4, 6, 8}.
Correct answer: D. [2, 5]
Explanation: In interval notation, a square bracket means that the endpoint is included, while a round bracket means that the endpoint is excluded. To include both 2 and 5, the interval must have a square bracket at both ends. Therefore [2, 5] is correct. The intervals (2, 5), [2, 5), and (2, 5] each exclude at least one endpoint, so none of them contains both numbers.
Correct answer: B. 3 is not included and 9 is included.
Explanation: The interval (3, 9] uses a round bracket at 3 and a square bracket at 9. The round bracket means that 3 is excluded, so 3 ∉ (3, 9]. The square bracket means that 9 is included, so 9 ∈ (3, 9]. Therefore statement B is correct. In general, an interval is interpreted by checking the bracket at each endpoint separately.
Correct answer: B. A = [0, 1]
Explanation: The condition 0 ≤ x ≤ 1 describes all real numbers from 0 through 1, including both endpoints. In interval notation, a square bracket means that the endpoint is included, while a round bracket means that it is excluded. Since both inequalities contain equality, both 0 and 1 belong to A. Therefore, A = [0, 1].
Correct answer: C. [-2, 2]
Explanation: The symbol R denotes the set of all real numbers, so x may be any real number satisfying the inequality. The condition -2 ≤ x ≤ 2 includes both endpoint values -2 and 2. In interval notation, an included endpoint is represented by a square bracket. Therefore, the set is written as [-2, 2]. It is not a finite roster set because infinitely many real numbers lie between -2 and 2.
Correct answer: A. (-2, 2)
Explanation: For a positive number 2, the inequality |x| < 2 means that the distance of x from zero is less than 2. Equivalently, -2 < x < 2. Both inequalities are strict, so neither -2 nor 2 is included; this requires round brackets at both ends. Therefore the correct interval notation is (-2, 2), not the outside region or a one-sided interval.
Correct answer: A. ∅ ∈ A
Explanation: The notation A = {∅, 1} means that A has exactly two elements: the empty set ∅ and the number 1. Therefore ∅ ∈ A is true. The set {1} is not an element of A, even though 1 itself is an element of A, so option B is false. No element belongs to ∅, making C false. Finally, {∅,1} equals A, so it is not a proper subset of A. Hence option A is correct.
Correct answer: A. {1, 3, 5, 15}
Explanation: The word “and” means that x must satisfy both conditions simultaneously. The positive divisors of 60 include 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Among these, the numbers that also divide 15 are 1, 3, 5, and 15. Therefore A = {1, 3, 5, 15}. Option B lists many divisors of 60 that do not divide 15, while option D omits 1 and 15.
Correct answer: A. {1, 2, 3, 4}
Explanation: Solve the inequality: 2x + 1 < 10 gives 2x < 9 and hence x < 4.5. Since x belongs to N = {1, 2, 3, ...}, the possible natural-number values are 1, 2, 3, and 4. The value 5 fails because 2(5)+1 = 11, and 0 is not in the stated definition of N. Therefore A = {1, 2, 3, 4}, option A.
Correct answer: B. 12
Explanation: The set \(A\) has four elements, so it has \(2^4=16\) total subsets. It is easier to count the complement: subsets containing neither 1 nor 2 can use only the elements 3 and 4, giving \(2^2=4\) subsets. Hence the number containing 1 or 2 is \(16-4=12\). Here “or” is inclusive, so subsets containing both 1 and 2 are also counted. Therefore, option B is correct.
